What is the formula for $$\int_0^{\pi } \frac{\sin (n \sigma )}{(a-\cos \sigma )^2} \, d\sigma$$
where $ a>1 $ and $n$ is a positive integer?
To evaluate, I tried to replace $a$ with $\cosh\xi $ in the integral
$$\int_0^{\pi } \frac{\sin (n \sigma )}{(\cosh\xi-\cos \sigma )^2} \, d\sigma$$
Please note that I already obtained the formula for a related integral $$\int_0^{\pi } \frac{\cos (n \sigma )}{(a-\cos \sigma )^2} \, d\sigma
= \frac{\pi(n \sqrt{a^2-1}+a)}{(a^2-1)^{3/2} (\sqrt{a^2-1}+a)^n}
$$
Best Answer
Evaluate \begin{align} K_n(r)= &\int_0^\pi \frac{e^{i n x}}{1-2r\cos x+r^2} \overset{z= e^{ix} }{dx} = \frac1i \int_\gamma\frac{z^n}{(z-r)( 1-rz)}dz\\ \end{align} along the semicircle path $\gamma$ in the upper plane, resulting in \begin{align} I_n(r) =& \int_0^\pi \frac{\cos (n x)}{1-2r \cos x +r^2}dx=\Re K_n(r) = \frac{\pi r^n}{1-r^2}\\ J_n(r)= &\int_0^\pi \frac{\sin (n x)}{1-2r \cos x +r^2}dx= \Im K_n(r)\\ =&\ \frac{{r^{-n}}-r^n} {1-r^2}\ln\frac{1+r}{1-r} -2\sum_{k=1}^{[\frac n2]}\frac{{r^{-n-1+2k}}-r^{n+1-2k}}{(2k-1)(1-r^2)} \end{align} Then, utilize $I_n(r)$ and $J_n(r)$ to evaluate \begin{align} & \int_0^\pi \frac{\cos (n x)}{\cosh\xi -\cos x }dx =2e^{-\xi} I_n(e^{-\xi})=\frac{\pi e^{-n\xi}}{\sinh \xi}\\ \\ &\int_0^\pi \frac{\sin( n x)}{\cosh\xi-\cos x }dx = 2e^{-\xi} J_n(e^{-\xi})\\ =& \ \frac{2\sinh(n\xi)}{\sinh \xi} \ln\left(\coth\frac\xi2\right) -4\sum_{k=1}^{[\frac n2]}\frac{\sinh(n+1-2k)\xi}{(2k-1)\sinh\xi} \end{align} which, via differentiation with respect to $\xi$, lead to \begin{align} & \int_0^\pi \frac{\cos (n x)}{(\cosh\xi -\cos x )^2}dx=\frac{\pi e^{-n\xi} (n+\coth\xi )}{\sinh^2 \xi}\\ \\ & \int_0^\pi \frac{\sin (n x)}{(\cosh\xi -\cos x )^2}dx\\ =& \ \frac{2}{\sinh^2 \xi}\bigg\{ \sin^2\frac{n\pi}2 + \left[\frac{\sinh(n\xi)}{\tanh\xi}-n\cosh(n\xi) \right]\ln\left(\coth\frac\xi2\right)\\ &\hspace{5mm} -\sum_{k=1}^{[\frac n2]}\frac{2}{2k-1} \left[\frac{\sinh(n+1-2k)\xi}{\tanh\xi} -n\cosh(n+1-2k)\xi\right]\bigg\} \end{align}